Question

Which are real zeroes of this function? x3 + 2x2 – 9x – 18
A. 1, 2, 3
B. 3, -3, 2
C. 2, 3, -6
D. 1, -1, 18
E. -2, 3, -3

Answer 1

Hello!

The answer is:

E. -2, 3, -3

Why?

To find the zeroes of the function, we just need to evaluate the given values and check if the function/expression tends to 0.

We are given the function

$f(x)=x^{3}+2x^{2}-9x-18$

Now, to find the zeroes, we need to substitute the values, so, substituting the values of the option E, we have:

First value, -2:

$f(-2)=-2^{3}+2*(-2)^{2}-9*(-2)-18$

$f(-2)=-8+2*4+18-18$

$f(-2)=-8+8+18-18=0$

Second value, 3:

$f(3)=3^{3}+2*(3)^{2}-9*(3)-18$

$f(3)=27+2*9-27-18$

$f(3)=27+18-27-18=0$

Third value, -3:

$f(-3)=-3^{3}+2*-3^{2}-9*-3-18$

$f(-3)=-27+2*9+27-18$

$f(-3)=-27+2*9+27-18=0$

Hence, we have that evaluating the values of the option E, the function tends to 0.

So, the correct option is: E. -2, 3, -3

Have a nice day!